[tex]f(x)=e^{2x-1}\\\\y=e^{2x-1}\\\\\text{replace x with y and y with x}\\\\x=e^{2y-1}\\\\\text{solve for y}\\\\e^{2y-1}=x\ \ \ \ |\ln\\\\\ln e^{2y-1}=\ln x\\\\2y-1=\ln x\ \ \ |+1\\\\2y=\ln x+1\ \ \ |:2\\\\y=\dfrac{\ln x+1}{2}\\\\Answer:\ \boxed{f^{-1}(x)=\dfrac{\ln x+1}{2},\ x > 0}[/tex]
